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The square root of sixteen is positive and negative four.

4b. Assume that f(3)=4. Name a point that must be on the graph of y= 1/2f(x/2).

The point (3, 4)  is on the originalgraph

The "x/2"  horizontally "stretches" the graph by a factor of 2

The (1/2)   vertically  "compresses" the graph by  a factor  of 1/2

So....a point on  y = (1/2)f(x/2)  must be     (2*3 , (1/2)*4)  =  (6, 2)

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Nirvana

4a. Assume that f(3)=4. Name a point that must be on the graph of y=f(x)+4.

The graph of  y = f(x) + 4    shifts  f(x) up by 4 units....so....

f(x) + 4    =   (3, 4 + 4)   =  ( 3 , 8)

3. Find all points (x,y) that are 13 units away from the point (2,7) and that lie on the line x-2y=10.

We can construct  a circle  centered at (2,7)  with a radius of 13

The equation of this circle is

(x - 2)^2 + (y - 7)^2  = 169       (1)

Rearrange the equation of the line  as  x = 2y+10     (2)

Put (2) int (1)  for  x     and we have

(2y + 10 - 2)^2 + ( y - 7)^2  = 169

(2y + 8)^2 + (y - 7)^2 = 169   simplify

4y^2  + 32y + 64  + y^2 - 14y + 49  = 169

5y^2 + 18y + 113 = 169       subtract 169 from both sides

5y^2 + 18y - 56   = 0     factor this

(5y  + 28) ( y - 2)  = 0

Set each factor to 0  and solve for y

5y + 28  = 0                     y  - 2  =  0

5y  = -28                           y  = 2

y  = -28/5

Put both y values back into the equation of the line to find their associated x coordinates

x - 2(-28/5)  = 10

x + 56/5  =  50/5

x = 50/5 - 56/5

x = -6/5

So one point  is  ( -6/5, -28/5)

And

x - 2(2)  = 10

x - 4  = 10

x = 14

And the other point is  (14, 2)

$10 AND $20 that equal $90 

1. 10x9=$90      so 9 10 dollar bills equals 90 

2. 4x20=80     1x10= 10    so  4 twenty dollar bills and 1 10 dollar bill equals $90 

3. 3x20=60    3x10= 30     so three twenty dollar bills and 3 ten dollar bills equals $90 

4. 2x20=40    5x10=50       so two twenty dollar bills and 5 ten dollar bills equals $90 

5. 1x20=20    7x10=70      so 1  twenty dollar bill and seven ten dollar bills equals $90 

1. What are the coordinates of the points where the graphs of f(x)=x^3-x^2+x+1 and g(x)=x^3+x^2+x-1 intersect?

Set these equal

x^3 - x^2 + x + 1  =  x^3 + x^2 + x - 1        subtract  x^3, x from both sides

-x^2  + 1 = x^2 - 1       add  x^2 to both sides....subtract 1 from both sides

0  = 2x^2 - 2

2x^2 - 2  = 0

x^2 - 1  = 0

x^2  = 1         take both roots

x = 1      or  x = -1

When x = 1 , y  =(1)^3 - (1)^2 + 1 + 1  =  2

So (1, 2)  is one intersection point

And when x = - 1 , y =  (-1)^3 - (-1)^2  - 1 + 1  =  -2

So  (-1,-2)  is the other intersection point

See the graph here : https://www.desmos.com/calculator/nnvu6b9qhr

EDIT TO CORRECT AN ERROR     

$90

9 tens

1 twenty and 7 tens

2 twenties  and 5 tens

3 twenties ans 3 tens

4 twenties and 1 ten

State that the function is odd, in which case the integral will be zero.

Notice that the integrand is an odd function i.e. if f(x) = sin(sinh(x)) then f(-x) = -f(x).  This means the integral of f(x) from -1 to +1 is zero. 

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